Super-Hard SAT Geometry Question 4 | Free SAT Prep

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Question 4·25 Super-Hard SAT Geometry and Trigonometry Questions·Geometry and Trigonometry

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A right square pyramid has a square base with side length $40$ inches. The pyramid's slant height, measured from the apex to the midpoint of a side of the base, is $29$ inches. A conical cavity is removed from the pyramid. The cone has the same apex as the pyramid, and its circular base is inscribed in the square base of the pyramid.

Which choice gives the volume, in cubic inches, of the material remaining in the pyramid?

For a cone removed from a pyramid when both share an apex and the same base plane, first determine their common vertical height. Then look for shared factors in the volume formulas: both a square pyramid and a cone have volume $\frac{1}{3}$ times base area times height. Subtract the base areas first, then multiply by the common factor to reduce arithmetic and avoid treating the cone as a cylinder.

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Use the slant height

Draw the right triangle formed by the apex, the center of the square base, and the midpoint of one side of the base.

Identify the cone's radius

Because the circular base is inscribed in the square, the circle's diameter equals the side length of the square.

Use the shared height

The pyramid and cone have the same vertical height. Factor out the common $\frac{1}{3}$ and height when subtracting their volumes.

Desmos Guide

Calculate the vertical height

The geometric setup is faster by hand. In the Desmos calculator, enter sqrt(29^2-20^2) to verify that the vertical height is $21$ .

Verify the remaining volume

Enter (1/3)(21)(40^2-pi(20^2)). Desmos will display a decimal value for the remaining volume, which corresponds to the exact expression in choice C.

Step-by-step Explanation

Find the vertical height

The apex, the center of the square base, and the midpoint of a base side form a right triangle. The distance from the center of the square to the midpoint of a side is half of $40$ , or $20$ .

\begin{aligned} h^2+20^2&=29^2\\ h^2&=441\\ h&=21 \end{aligned}

Find the two base areas

The square base has area $40^2=1600$ .

The circle is inscribed in the square, so its radius is $20$ . Its area is $\pi(20)^2=400\pi$ .

Subtract the volumes

Both solids have height $21$ , and each has volume equal to $\frac{1}{3}$ times its base area times its height.

\begin{aligned} V_{\text{remaining}}&=\frac{1}{3}(21)(1600-400\pi)\\ &=11200-2800\pi \end{aligned}

Thus, choice C is correct.

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